# Thread: Equliateral Triangle inscribed in circle

1. ## Equliateral Triangle inscribed in circle

Equilateral triangle ABC is inscribed in a circle. The angle bisector at A intersects the circle again at P (AP is the diameter). If the length of segment AC is 10, find AP - (BP + CP).

I will upload a picture soon, but help is greatly appreciated!!

2. Originally Posted by hachiko
Equilateral triangle ABC is inscribed in a circle. The angle bisector at A intersects the circle again at P (AP is the diameter). If the length of segment AC is 10, find AP - (BP + CP).
Let O be the centre of the circle. Can you see why OP = BP = CP? What does that tell you about AP – (BP + CP)?

3. Hello, hachiko!

Another approach . . .

Equilateral triangle $ABC$ is inscribed in a circle.
The angle bisector at $A$ intersects the circle again at $P$ ( $AP$ is adiameter).
If the length of segment $AC$ is 10, find: . $AP - (BP + CP)$
Code:
                A
* o *
*    /|\    *
*     / | \     *
*     /  |  \ 10  *
/   |30°\
*    /    |    \    *
*   /     |     \   *
*  /      |      \  *
/       |       \
B o- - - - + - - - -o C
*       |     * *
* 60° |  *  *
* o *
P

We have: inscribed equilateral triangle $ABC\!:\;\;AC = 10$
. . with diameter $AP,\;\;\angle CAP = 30^o.$
Draw chord $CP$.

$\Delta ACP$ is inscribed in a semicircle.
. . Hence: . $\angle ACP = 90^o,\;\;\angle APC = 60^o$

We have: . $CP = x,\;\;AC = 10,\;\;AP = 2x$ .(30°-60° right triangle)

Pythagorus says: . $CP^2 + AC^2 \:=\:AP^2 \quad\Rightarrow\quad x^2 + 10^2 \:=\:(2x)^2$

. . $x^2 + 100 \:=\:4x^2 \quad\Rightarrow\quad 3x^2 \:=\:100 \quad\Rightarrow\quad x^2 \:=\:\frac{100}{3}\quad\Rightarrow\quad x \:=\:\frac{10\sqrt{3}}{3}$

Hence: . $BP \,=\,CP \,=\,\frac{10\sqrt{3}}{3},\;\;AP \,=\,\frac{20\sqrt{3}}{3}$

Therefore: . $AP - (BP + CP) \;\;=\;\;\frac{20\sqrt{3}}{3} - \left(\frac{10\sqrt{3}}{3} + \frac{10\sqrt{3}}{3}\right) \;\;=\;\;0$